Question 12.17: A straight conical draft tube having an inlet diameter of 2.......
A straight conical draft tube having an inlet diameter of 2.75 m is connected at the exit of a propeller turbine operating under a head of 7 m and developing 1700 kW power with an overall efficiency of 88%. The turbine is set 2.3 m above the tail race level and the vacuum gauge inserted at the turbine outlet records a suction head of 3.2 m. Calculate the efficiency of the draft tube, if the loss of head due to friction in the draft tube is 0.25 of the kinetic head at outlet.
Given: d_2=2.75 m ; H=7 m ; S P=1700 \times 10^3 W ; \eta_o=0.88 ; z_2=2.3 ; \left(p_a-p_2\right) / w =3.2 m ; h_f=0.25 V_3^2 / 2 g
Discharge may be calculated from the equation, \eta_o=S P / w Q H
Discharge Q=\frac{S P}{w H \eta_o}=\frac{1700 \times 10^3}{9810 \times 7 \times 0.88}=28.132 m ^3 / s
V_2=\frac{Q}{A_2}=\frac{4 Q}{\pi d_2^2}=\frac{4 \times 28.132}{\pi \times 2.75^2}=4.74
and h_f=\frac{0.25 V_e^2}{2 g}
∴ \frac{p_2}{W}-\frac{p_e}{W}=-H_s-\left(\frac{4.74^2-V_3^2}{2 g}-\frac{0.25 V_3^2}{2 g}\right)
-3.2=-2.3-\left(\frac{4.74^2}{2 g}-\frac{1.25 V_3^2}{2 g}\right)
∴ V_3=1.955 m / s
\begin{aligned}\frac{V_2^2-V_3^2}{2 g}&=\frac{4.74^2-1.955^2}{2 \times 9.81}=0.9504 \\\frac{V_2^2}{2 g}&=\frac{4.74^2}{2 \times 9.81}=1.14 \\h_f &=0.25 \frac{V_3^2}{2 g}=\frac{0.25 \times 1.955^2}{2 \times 9.81}=0.0487\end{aligned}
Efficiency of draft tube \eta_d=\frac{0.9504-0.0487}{1.145}=0.7875=78.75 \%